Q&A - PSLE Math

[tianzhu]

Hi Rockstar and VC


Good Morning

Thank you, I am sure PSLE students who are reading this thread will benefit from your solutions.

I picked up these three questions from OnSponge, they were from a P5 member who is preparing for her Maths Olympiad exam.I felt it would be good for KSP's PSLE kids and their parents to take a look at them.

Do you find Q2 familiar? Questions of such genre have been found frequently in past years papers.

Ann is driving to a destination. If she travels at 80km/h, she would be 20min late. If she travels at 60km/h, she would be 45 min late. How long will journey take if she travels at 90km/h.

Q3 is quite similar to PSLE 2009 Speed question

Mei and Lin were in a bicycle race. Mei was travelling at a constant speed of 20km/hr and they both did not change their speed. When Lin completed half the race, Mei was 3.5km ahead. Mei completed the race at 10.45am. What time did Lin complete the race? Answer: 11.06am

Best wishes

[Dharma]

Hi,


Below are 3 interesting speed questions posted by Tianzhu and I believe that these are some of the more challenging ones that a child preparing for PSLE must learn how to handle.

Besides knowing the distance, speed and time [DST] relationship which is of utmost importance to solve speed related questions, what is important and crucial is the ability to look at the question and try to find the key information in the question and knowing how to use the DST relationship in the correct manner in order to solve the question easily.

If we are able to handle the above, speed questions may not be as difficult as they seem.

Q1
Last Monday, Danny walked from home to market. For the first 5 minutes, he was walking at an average speed of 35m/min. When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes. Find the distance between market and home.

Key information :
When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes.

Question tells us that with different speeds, Danny will take different times to reach the market. In order to use ratio method of the DST relationship, we need to make distance fixed/constant.

We consider the distance taken by Danny from the point he was 5 minutes after he started from home to the market. (Distance is fixed)

Speed ratio => 7 : 9
Time ratio => 9 : 7
9u – 7u = 2u = 4mins + 2 mins = 6 mins
1u = 3 mins
9u = 27 mins
Total time from home to market = 5mins + 27 mins = 32 mins
Distance = 35m/min x 32 mins = 1120m

Q2
Jacki’s mum leaves home at 12.35pm every day and arrives 25 minutes later at his school to pick him up. She always cycles at the same speed along the same route. One day Jacki was dismissed early at 12.30pm, so he started walking home at 6km/h, hoping to meet her along the way. He met her at x and got onto her bicycle. They reached home 15 minutes earlier than before. How far did Jacki walked before his mum picked him up?

Key information :
They reached home 15 minutes earlier than before

If they reached home 15 mins earlier, it means that 15 mins was saved by Jacki’s mum to travel from the point she met Jacki to school and back to the same point. That means her would have met Jacki 7.5 mins before 1.00pm ( Mum left home at 12.35pm and takes 25 mins normally to reach school).
So, since Jacki’s left school at 12.30pm, he will take 22.5 mins before he meets his mum.

Distance Jacki walked = 22.5/60 hrs x 6km/h = 3/8hrs x 6km/hr = 2.25km

Q3
Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles reach the finishing line?

Key information :
Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him.

When Charles covered 1/3 of race; Tony was 15km in front.
When Charles covered 3/3 of race; Tony will be 45km in front.

Time taken by Tony to travel 45km = 45km/ 90km/hr = ½ hr
So, Charles reached the finishing line ½ hr after Tony.
Charles reached the finishing line at 10.05 a.m.

[pecalis]

Dharma:

Hi,


Below are 3 interesting speed questions posted by Tianzhu and I believe that these are some of the more challenging ones that a child preparing for PSLE must learn how to handle.

Besides knowing the distance, speed and time [DST] relationship which is of utmost importance to solve speed related questions, what is important and crucial is the ability to look at the question and try to find the key information in the question and knowing how to use the DST relationship in the correct manner in order to solve the question easily.

If we are able to handle the above, speed questions may not be as difficult as they seem.

Q1
Last Monday, Danny walked from home to market. For the first 5 minutes, he was walking at an average speed of 35m/min. When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes. Find the distance between market and home.

Key information :
When he realized that he was going` to be late by 4 minutes, he quickly increased his speed by 10m/min. As a result, he was early by 2 minutes.

Question tells us that with different speeds, Danny will take different times to reach the market. In order to use ratio method of the DST relationship, we need to make distance fixed/constant.

We consider the distance taken by Danny from the point he was 5 minutes after he started from home to the market. (Distance is fixed)

Speed ratio => 7 : 9
Time ratio => 9 : 7
9u – 7u = 2u = 4mins + 2 mins = 6 mins
1u = 3 mins
9u = 27 mins
Total time from home to market = 5mins + 27 mins = 32 mins
Distance = 35m/min x 32 mins = 1120m

Q2
Jacki’s mum leaves home at 12.35pm every day and arrives 25 minutes later at his school to pick him up. She always cycles at the same speed along the same route. One day Jacki was dismissed early at 12.30pm, so he started walking home at 6km/h, hoping to meet her along the way. He met her at x and got onto her bicycle. They reached home 15 minutes earlier than before. How far did Jacki walked before his mum picked him up?

Key information :
They reached home 15 minutes earlier than before

If they reached home 15 mins earlier, it means that 15 mins was saved by Jacki’s mum to travel from the point she met Jacki to school and back to the same point. That means her would have met Jacki 7.5 mins before 1.00pm ( Mum left home at 12.35pm and takes 25 mins normally to reach school).
So, since Jacki’s left school at 12.30pm, he will take 22.5 mins before he meets his mum.

Distance Jacki walked = 22.5/60 hrs x 6km/h = 3/8hrs x 6km/hr = 2.25km

Q3
Tony and Charles took part in a car race. Tony drove at a speed of 90km/h. Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him. Tony reached the finishing line at 9.35 a.m. At what time did Charles reach the finishing line?

Key information :
Both of them did not change their speed throughout the race. When Charles had covered 1/3 the distance, Tony was 15km in front of him.

When Charles covered 1/3 of race; Tony was 15km in front.
When Charles covered 3/3 of race; Tony will be 45km in front.

Time taken by Tony to travel 45km = 45km/ 90km/hr = ½ hr
So, Charles reached the finishing line ½ hr after Tony.
Charles reached the finishing line at 10.05 a.m.

Thanks Dharma,
For sharing the solutions and the clear explanation. Really appreciate it

[atutor2001]

Vanilla Cake:

leesf:

Q3) A street-lamp is erected at an interval of 8m on one side of the street. On the other side of the street, an angsana tree is planted at every 6m. There are 12 more angsana trees than street-lamps.

a) Ho long is the street if there is one street-lamp and one angsana tree on both ends of the street.
b) How many street-lamps and angsana trees are there?

atutor2001:

Q3 The 1st common multiple of 6 & 8 is 24 (i.e. 3 gaps between 2 lamp and 4 gaps between 2 trees)
4 - 3 = 1
Therefore, for there to be 12 more trees than lamp we need to repeat this interval 12 times i.e. the 12 common multiples of 6 & 8

(a) Street length = 12 x 24 = 288m
(b) No. of trees = 5 + 11 x 4 = 49 No. of lamps = 4 + 11 x 3 = 37

(after the first group, the no. of trees & lamps is reduced by 1 because there is already a \"reference tree/lamp\" from the end of the last group\"

e.g.
x x x x)x x x)x x x) - x represents lamps
|--24-|--24-|--24-|[/u]

Lowest common multiple of 6 and 8 is 24.
24÷6=4
4+1=5
So, there are 5 angsana trees for the first 24m.

24÷8=3
3+1=4
So, there are 4 street-lamps for the first 24m.

From the above workings, you can see that there is 1 additional angsana tree for every 24m.
Since there are 12 more angsana trees than street-lamps (given in the question itself), length of the street -> 24÷1x12=288m
(a) If there is one street-lamp and one angsana tree on both ends of the street, length of street = 288m

288÷8=36
36+1=37 street-lamps.
Number of angsana trees -> 37+12=49.
(b) Numbers of street-lamps and angsana trees are 37 and 49 respectively.

Hi Vanilla Cake

Thank you for correcting my mistake. Got confused between trees and distances. Your method for finding the number of trees and lamps is so much better.

I have amended my working in red so as not to cause further confusion. Apology to anyone who got confused by me.

[tianzhu]

leesf:

Q3) A street-lamp is erected at an interval of 8m on one side of the street. On the other side of the street, an angsana tree is planted at every 6m. There are 12 more angsana trees than street-lamps.
a) Ho long is the street if there is one street-lamp and one angsana tree on both ends of the street.
b) How many street-lamps and angsana trees are there?

Hi

This question is quite similar to the balloon question in PSLE Maths 2009.

String of 2 big balloons is 90cm
String of 5 small balloons is is 1.2m
If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether?

Best wishes

[Dharma]

Q3) A street-lamp is erected at an interval of 8m on one side of the street. On the other side of the street, an angsana tree is planted at every 6m. There are 12 more angsana trees than street-lamps.

a) How long is the street if there is one street-lamp and one angsana tree on both ends of the street.
b) How many street-lamps and angsana trees are there?

Spacing ratio => Street lamp : Angsana Tree = 4u : 3u = 4 : 3
Qty ratio => Street lamp : Angsana Tree = 3u + 1 : 4u + 1

(4u + 1 ) – (3u + 1 ) = 1u = 12
Length of street = (3x12) x 8m = 288m or (4 x 12) x 6 = 288m

No. of street lamps = (3 x 12) + 1= 37
No. of angsana trees = (4 x 12) + 1 = 49

[Dharma]

pecalis:

Thanks Dharma,
For sharing the solutions and the clear explanation. Really appreciate it

Glad u like it.

[tianzhu]

Hi


Hope this helps.

Best wishes.

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[octopusmum]

Hi Vanilla cake

thks for telling me the solutions for the 2 maths questions. i simply did not have the time to search for the answers myself so have to trouble kind souls like you to help…

i need help again with the following question from Henry Park maths 2010 prelim paper. it is qtn 11 in paper 2:

The figure shows an empty tank. It is made from 2 containers. The containers are in the form of 2 cubes of sides 8 cm and 20 cm respectively. The small container is attached to the centre of one of the sides of the big container. 4 litres of water is poured into the big container such that water flows in to fill part of the small container. What is the height of the water level in the big container? Round off your answer to 2 decimal places.

it is easier to see the question with the diagram but i am not able to reproduce the diagram here…if you or anyone else happen to know the solution or know where to get the detailed solution, pls share.

thks v much

a very desperate and exhausted octopusmum

[kancheongmum]

octopusmum:

Hi Vanilla cake

thks for telling me the solutions for the 2 maths questions. i simply did not have the time to search for the answers myself so have to trouble kind souls like you to help...

i need help again with the following question from Henry Park maths 2010 prelim paper. it is qtn 11 in paper 2:

The figure shows an empty tank. It is made from 2 containers. The containers are in the form of 2 cubes of sides 8 cm and 20 cm respectively. The small container is attached to the centre of one of the sides of the big container. 4 litres of water is poured into the big container such that water flows in to fill part of the small container. What is the height of the water level in the big container? Round off your answer to 2 decimal places.

it is easier to see the question with the diagram but i am not able to reproduce the diagram here..if you or anyone else happen to know the solution or know where to get the detailed solution, pls share.

thks v much

a very desperate and exhausted octopusmum

Hi octopusmum

I take the liberty to help you with this question as DS also need help.
The basic concept is volume = L x B x H
what is L x B? Is the base area right. So keep this in mind
Total volume water 4L = 4000 cm3
volume of water below 8 cm cube = 20 x 20 x 6 ((20 - 8/2))=2400cm3
the remaining water above the 8cm cube will have different base area.
= 4000 - 2400 = 1600cm3
new base area is (20x20) + (8x8)= 464 cm2
vol of remaining water divide by this new base area you will get the height. 1600/464 = 3.45 cm
3.45 + 6 = 9.45 cm

Nanyang prelim 2010 Q11 is almost the same as above question. You may want to let your child try.

Let us know if you need help.